# Tag Info

2

Note the projections $\pi_1 : \mathbb R^n \to \mathbb R^m$ and $\pi_2 : \mathbb R^n \to \mathbb R^{n-m}$ defined by $$\pi_1(x_1, \ldots, x_n) = (x_1, \ldots, x_m), \quad \text{and} \quad \pi_2(x_1, \ldots, x_n) = (x_{m+1}, \ldots, x_n)$$ are continuous. Then $g \star h = (g \circ \pi_1) (h \circ \pi_2)$ is continuous as compositions and products of ...

1

$\int f<\infty$ can only be true if $f$ is integrable, i.e. if $f$ is measurable and $\int |f|<\infty$. If $|f|\leq M$ then $f^2\leq M|f|$ so that: $$\int f^2\leq\int M|f|=M\int|f|<\infty$$

3

It is given that $f \geq 0$. $\int f^{2}d\mu \leq M\int fd\mu<\infty$ (where $M$ is such that $f(x) \leq M$ for all $x$).

3

Just look at the terms on the right side. The terms $-\frac 1 {|B|} \int_B g(y)dy$ and $+\frac 1 {|B|} \int_B g(y)dy$ cancel each other and so do the terms $-g(x)$ and $+g(x)$.

4

Lebesgue's differentiation theorem does hold for a much larger class of measures; it is not restricted to Lebesgue measure. The following statement is compiled from Measure Theory Vol. 1 by Bogachev (Theorem 5.8.8). Let $\mu$ be a measure on $(\mathbb{R}^n,\mathcal{B}(\mathbb{R}^n))$ which is finite on all balls. If $f \in L^1(\mu)$, then $$f(x) = \lim_{r ... 0 You don't need DCT for this. Use the fact that |\frac {\sin t} {1+t^{2}}| \leq \frac 1 {t^{2}}I_{|t|>1}+I_{|t|\leq 1}. 2 Your idea using DCT works. To show that F is continuous in any x_0\in\mathbb{R}, consider a sequence (x_n) with x_n\rightarrow x_0. Then the sequence of functions h_n(t):=f(t)\,\chi_{(-\infty,x_n)}(t) converges pointwise a.e. to h(t):=f(t)\,\chi_{(-\infty,x_0)}(t), and |h_n(t)|\leq f(t). Since f is integrable, DCT implies$$ \lim_{n\...

2

It is false. Choose any $b>a>0$. Define $q$ as $$q(x)=\begin{cases}0,x\in(-\infty,0)\cup[2,\infty)\\a,x\in[0,1)\\b,x\in[1,2)\end{cases}$$ Take $\phi$ as any function in $C^1_c(\mathbb{R})$ that is $1$ on $[0,2]$ and $0$ outside $(-1,3)$. Assuming the conjecture is true, we should have \lim_{\eta\rightarrow 0}\frac{1}{\eta}\int_{\mathbb{R}}(q(x+\... 0 n\int_{A_n} g_n \to 0 for every uniformly bounded sequence \{g_n\} iff n \mu \{x: f_n(x) \geq \frac 1 n\} \to 0. [In one direction take g_n=1 for all n to prove this]. 2 Integration and differential calculus are closely releted topics and there is no reason why one should use row vectors in one and column vectors in the other. The choice between row vectors in one and column vectors is a personal one and one can use either one in any topic. 4 Suppose x_n is a sequence in (0,1) where |g(x_n)| > 2^n. Define a function f on (0,1) so that f(x) = x_n on (2^{-n-1}, 2^{-n}]. 2 Your solution seems fine. For a bit of further simplification, we may do as follows. Write I_n for the integral. Then \begin{align*} &\left| I_n - \int_{1}^{\infty} \frac{\cos(n\pi x)}{x^{2}} \, \mathrm{d}x \right| \\ &\leq \int_{0}^{1} \left| \frac{x^{n-2}\cos(n\pi x)}{1+x^n}\right| \, \mathrm{d}x + \int_{1}^{\infty} \left| \frac{x^{n-2}}{1+x^n} -... 1 Hint: If x \geq 1 then \frac{x^{n-2}}{1+x^n} \leq \frac{1}{x^2}$$and if 0\leq x \leq 1 then$$ \frac{x^{n-2}}{1+x^n} \leq x^{n-2} \leq 1 $$3 DCT requires the condition that the integrand is dominated by an integarble function. In this case if \frac 1 n I_{[n, \infty)} \leq g then g \geq \frac 1 n on the interval [n,\infty) for all n and, in particular for n=1, so \int_0^{\infty} g =\infty. Thus there is no dominating integrable function,. 1 Hint: apply Fatou's lemma to the sequence of functions f_k = f 1_{E_k}. 3 Hints: Since the intervals [n,n+1) are disjoint, it holds that$$|f| = \sum_{n =1}^{\infty} |a_n| \chi_{[n,n+1)}.$$Conclude that$$\int |f| = \sum_{n=1}^{\infty} |a_n|$$(e.g. using monotone convergence). This gives you a sufficient and necessary condition for integrability of f. To show that convergence of \sum_{n=1}^{\infty} a_n is not sufficient, ... 2 No. Let E=(0,1) and f(x)= \frac{1}{\sqrt{x}}. 4 Consider n = 1, \Omega = (0,1) and let$$u_n(x) = \begin{cases} 1 \qquad k2^{-n} \leq x < (k+1)2^{-n} \text{ for an even } k\\ -1 \qquad \text{otherwise} \end{cases}Then you can check that u_n \rightharpoonup 0 weakly in L^2(\Omega) but |u_n(x)| = 1 for every x so \int |u_n| \phi \not \to 0 for \phi such that \int \phi \neq 0. To ... 3 This is related to the "duality characterization" of weak L^2- space. For a given \lambda>0 (which will be chosen later), let \displaystyle E = \left(E\cap \{|f|>\lambda\}\right) \cup \left( E\cap \{|f| \le \lambda\}\right) = E' \cup E'' . Then we have \begin{align*} \left|\int_E f\right| \le& \int_{E'} |f| + \int_{E''}|f| \\ \le& \int_\... 3 Yes, surely. 0 \leq \mu (A_i \cap E) \leq \mu (E) implies that \mu (A_i \cap E) =0. 3 Yes, by \sigma-additivity, \mu(E)=\mu(A_i \cap E)+\mu(E\setminus A_i) and, being a measure, \mu(E\setminus A_i)\geq 0. 0 I think it is possible to write your procedure rigurously in the following way. First of all notice that we can re-write w(t,x) as w(t,x)=\dfrac{u-\vert u\vert}{2}. $$Thus, differentiating we have$$ w_t=\dfrac{1}{2}\big(u_t-\hbox{sgn}(u)u_t\big)=\dfrac{1}{2}(1-\hbox{sgn}(u))u_t. $$Now, replacing this in E'(t) we obtain$$ E'(t)=2\int w w_t=\int (\...

2

The Tonelli and Fubini theorems are two of the most important results from measure theory to allow multi-dimensional integrals to be calculated over sets that aren't suitably elementary. Furthermore, the Lesbegue Dominated Convergence Theorem shows up all the time to calculate limits and derivatives of intgrals.

1

Take $E=[0,1]$ and, for each $x\in\mathbb R$, take $y=x-\lfloor x\rfloor$. Then $x-y\in\mathbb Z\subset\mathbb Q$.

2

Hint: given any interval $I_0$, find a nested sequence of intervals $I_0 \supset I_1 \supset I_2$ such that, for every $j\ge0$, there exists $n(j)$ such that $$I_j \subset \bigg[ q_{n(j)}-\frac1{{n(j)}^2},q_{n(j)}+\frac1{{n(j)}^2} \bigg].$$ The density of the rational numbers in $\Bbb R$ should help you construct such intervals. Do you see why we have $f(x)... 3 It's pretty hard to give a hint which will not get you to the answer. Well, by the definition of Lebesgue integration we know that the function$xf^2(x)$is integrable on both intervals$[0,\infty)$and$(-\infty,0)\$. Alright, so we have: $$\int_{[0,\infty)}xf^2(x)d\mu\geq \int_{M_y\cap[0,\infty)} xf^2(x)d\mu\geq y^2\int_{M_y\cap [0,\infty)}xd\mu$$ Now do ...

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